A fence 6 feet tall runs parallel to a tall building at a distance of 5 feet from the building. What is the length of the shortest ladder that will reach from
the ground over the fence to the wall of the building?
feet

Question

UpStudy AI · Accepted Answer

Let the building lie along the vertical line $$ x=0 $$ and the ground along the horizontal line $$ y=0 $$. The fence is parallel to the building and is located at $$ x=5 $$ with a height of 6 feet; hence its top is the point $$ (5,6) $$.

Suppose the ladder rests on the ground at a point $$ (d,0) $$ with $$ d>5 $$ (so that the foot is on the open side of the fence) and leans against the building, touching it at some point $$ (0,h) $$ where $$ h>6 $$. The ladder is a straight segment joining $$ (d,0) $$ and $$ (0,h) $$. For the ladder to just go over the fence it must pass through the point $$ (5,6) $$.

**Step 1.** Write the equation of the line through $$ (d,0) $$ and $$ (0,h) $$.

The slope of this line is 
$$
m=\frac{h-0}{0-d}=-\frac{h}{d}.
$$
Thus, the equation of the line is
$$
y = -\frac{h}{d}(x-d).
$$

At $$ x=5 $$ (the fence), the height is
$$
6 = -\frac{h}{d}(5-d).
$$
Notice that $$ 5-d $$ is negative (since $$ d>5 $$); multiplying both sides by $$-1$$ we have
$$
6 = \frac{h}{d}(d-5).
$$
Solve for $$ h $$:
$$
h = \frac{6d}{d-5}.
$$

**Step 2.** Express the ladder’s length.

The length of the ladder is the distance between $$ (d,0) $$ and $$ (0,h) $$:
$$
L = \sqrt{d^2 + h^2} = \sqrt{d^2 + \left(\frac{6d}{d-5}\right)^2}.
$$
For minimization, it is easier to work with the square of the length:
$$
F(d)=L^2=d^2+\frac{36d^2}{(d-5)^2}.
$$

**Step 3.** Differentiate $$ F(d) $$ with respect to $$ d $$ and find its critical point.

Differentiate term‐by‐term. The derivative of $$ d^2 $$ is $$ 2d $$. For the second term, write
$$
G(d)=\frac{36d^2}{(d-5)^2}.
$$
Using the quotient rule with 
$$
u(d)=d^{2},\quad v(d)=(d-5)^2,
$$
we have $$ u'(d)=2d $$ and $$ v'(d)=2(d-5) $$. Then,
$$
G'(d)=36\,\frac{u'(d)v(d)-u(d)v'(d)}{v(d)^2} 
=36\,\frac{2d(d-5)^2- d^2\cdot 2(d-5)}{(d-5)^4}.
$$
Factor out common terms in the numerator:
$$
2d(d-5)^2-2d^2(d-5)=2d(d-5)\left[(d-5)-d\right]=2d(d-5)(-5)=-10d(d-5).
$$
Then,
$$
G'(d)=36\, \frac{-10d(d-5)}{(d-5)^4}=-\frac{360d}{(d-5)^3}.
$$

So the derivative of $$ F(d) $$ is
$$
F'(d)=2d-\frac{360d}{(d-5)^3}.
$$
We set $$ F'(d)=0 $$ (and $$ d\neq 0 $$):
$$
2d-\frac{360d}{(d-5)^3}=0.
$$
Cancel $$ d $$ (since $$ d>0 $$):
$$
2-\frac{360}{(d-5)^3}=0 \quad\Longrightarrow\quad \frac{360}{(d-5)^3}=2.
$$
Solve for $$ d-5 $$:
$$
(d-5)^3=\frac{360}{2}=180 \quad\Longrightarrow\quad d-5=180^{1/3}.
$$
Thus,
$$
d=5+180^{1/3}.
$$

**Step 4.** Find $$ h $$ and then the ladder length $$ L $$.

Recall that
$$
h=\frac{6d}{d-5}=\frac{6\bigl(5+180^{1/3}\bigr)}{180^{1/3}}.
$$

Now, the ladder’s length is
$$
L=\sqrt{d^2+h^2} 
=\sqrt{\Bigl(5+180^{1/3}\Bigr)^2 + \left(\frac{6\bigl(5+180^{1/3}\bigr)}{180^{1/3}}\right)^2}.
$$
Factor out $$ 5+180^{1/3} $$:
$$
L=\bigl(5+180^{1/3}\bigr)\sqrt{1+\frac{36}{\left(180^{1/3}\right)^2}}.
$$

**Step 5.** Compute a numerical approximation.

First, approximate the
The shortest ladder length is approximately 10.95 feet.

A fence 6 feet tall runs parallel to a tall building at a distance of 5 feet from the building. What is the length of the shortest ladder that will reach from
the ground over the fence to the wall of the building?
feet

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A fence 6 feet tall runs parallel to a tall building at a distance of 5 feet from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building? feet

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A fence 6 feet tall runs parallel to a tall building at a distance of 5 feet from the building. What is the length of the shortest ladder that will reach from
the ground over the fence to the wall of the building?
feet